1. Ch. 17 Electrochemistry Ch. 17.3-17.4, 17.7

2. The relationship between the thermodynamics and electrochemistry. The work that can be accomplished when electrons are transferred through a wire depends on the “push” (the thermodynamic driving force). 17.3 Cell Potential, Electrical Work, and Free Energy

3. This driving force (the emf) a potential difference (in volts) between two points in the circuit. emf = potential difference (V) = work(J) charge(C) 2

4. 1 joule of work is produced or required when 1 coulomb of charge is transferred between two points that differ by a potential of 1 volt. Work is viewed from the point of view of the system. 3

5. Work flowing out of the system by a minus sign. When a cell produces a current, the cell potential is positive, the current can be used to do work. 4

6. Thus the cell potential E and the work w have opposite signs: E = -w ← Work q ← Charge -w = qE Maximum cell potential:

7. ● to obtain electrical work, current must be flowing  Some energy is wasted through frictional heating, the maximum work is not obtained ● In any real, spontaneous process some energy is always wasted-the actual work realized is always less than the calculated maximum.

8. Suppose a certain galvanic cell has a maximum potential (at zero current) of 2.50 V. 1.33 moles of electrons were passed through this cell at an average actual potential of 2.10 V. The actual work done is w=-qE E - the actual potential difference (2.10 V or 2.10 J/C q- the quantity of charge in coulombs transferred

9. The charge on 1 mole of electrons: a constant called the faraday (abbreviated F). 96.485 coulombs of charge per mole of electrons q = nF = 1.3 mol e x 96.485 C/mol e - w=-qE = -(1.33 mol e x 96.485 C/mol e - ) x (2.10 J/C) = -2.69 x 10 5 J

10. ● The change in free energy equals the maximum useful work obtainable from that process: w max = △ G w max = -qE max = △ G q = nF △ G = -qE max = -nFE max Potential of a galvanic cell to free energy

11. △ G = -nFE For standard conditions. △ G o = -nFE o The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell.

12. Calculate △ G o Cu 2+ (aq) + Fe (s) → Cu (s) + Fe 2+ (aq) Is this reaction spontaneous? △ G o = -nFE o Ex 17.3 Calculating △ G o for a Cell reaction

13. Two electrons are transferred per atom n = 2 mol e - F = 96.485 C/mol e- E o = 0.78 V = 0.78 J/C The process is spontaneous, the negative sign of △ G o, the positive sign of E o cell Cu 2+ (aq) + Fe (s) → Cu (s) + Fe 2+ (aq) △ G o = -nFE o

14. ● Predict whether 1 M HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution ● Write the half-reaction for HNO 3 (an oxidizing agent) NO 3 - + 4H + + 3e - → NO + 2H 2 O E o (cathode) = 0.96 V Oxidation of solid gold to Au 3+ ions Au → Au 3+ + 3e - -E o (anode) = -1.50 V ● E o value is negative. Ex. 17.4 Predicting Spontaneity

15. ● Process will not occur under standard conditions ● FYI: A mixture (1:3 by volume) of concentrated nitric and hydrochloric acids, called aqua regia, is required to dissolve gold.

16. The dependence of the cell potential on concentration. Under standard conditions (all concentrations 1 M). Cu(s) + 2Ce 4+ (aq) → Cu 2+ (aq) + Ce 3+ (aq) 17.4 Dependence of Cell Potential on Concentration

17. Potential of 1.36 V if [Ce 4+ ] is greater than 1.0 M? ● An increase in the concentration of Ce 4+ = forward reaction, increase the driving force on the electrons ● Cell potential will increase Cu(s) + 2Ce 4+ (aq) → Cu 2+ (aq) + Ce 3+ (aq)

18. Potential of 1.36 V if [Ce 4+ ] is greater than 1.0 M? ● increase in the concentration of a product (Cu 2 + or Ce 3+ )  decreasing the cell potential Cu(s) + 2Ce 4+ (aq) → Cu 2+ (aq) + Ce 3+ (aq)

19. For the cell reaction 2Al(s) + 3Mn 2+ (aq) → 2Al 3+ (aq) + 3Mn(s) E o cell = 0.48 V predict whether E cell is larger or smaller than E o cell for the following cases: a.[Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M b.[Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M 17.5The Effects of Concentration of E

20. E o cell = 0.48 V a.[Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M a.Increase the amount of products, then the reaction favors reverse. E cell <0.48V b.[Al 3+ ] = 1.0 M, [Mn 2+ ] = 3.0 M a.Increase the amount of reactant, then the reaction favors forward. E cell >0.48V 2Al(s) + 3Mn 2+ (aq) → 2Al 3+ (aq) + 3Mn(s)

21. Same compartments at different concentrations, we can construct galvanic cells. both contain AgNO 3 with different molarities. Concentration Cells

22. the direction of electron flow? Ag + + e - → Ag E = 0.80 V If cell had 1 M Ag + E cell = 0.80 V-0.80V = 0 V [Ag + ] are equal, electrons don’t flow Concentration Cells

23. [Ag + ] are unequal, half-cell potentials will not be identical The cell will exhibit a positive voltage Concentration Cells 0.1M and 1M

24. nature will try to equalize the concentrations of Ag + by transferring electrons from 0.1 M Ag + to 1 M and consume Ag + (to form Ag). Concentration Cells 0.1M and 1M

25. Electrons flow? Try to equalize the concentrations of Ag+ by transferring e-s from 0.1 M Ag + to 1 M (Left) Ag  Ag + + e- (increase Ag + ) (Right) Ag + +e-  Ag (decrease Ag + )

26. Both compartments have the same components but at different concentrations The difference in concentration is the only factor that produces a cell potential and the voltages are typically small. Concentration Cells

28. Warm up 3/24/15 1.Anode/cathode 2.The direction of electron flow

29. the direction of electron flow from the left to the right. Fe 2+ to be formed in the left and iron metal will be deposited on the right electrode. Ex. 17.6 Concentration Cells

30. the anode and cathode Oxidation occurs in the left (the anode) reduction occurs in the right (the cathode) Ex. 17.6 Concentration Cells

31. Recall this from ch. 16 ∆ G = ∆ G o + RTln(Q) (Q is the reactant quotient) -nFE = -nFE o + RTln(Q) E = E o - (RT/nF)ln(Q) The cell potential and the concentration of the cell at 25 C E = E o - (0.0592/n)log(Q) (The Nernst Equation) The Nernst Equation

32. 2Al(s) + 3Mn 2+ (aq) → 2Al 3+ (aq) + 3Mn(s) [Mn 2+ ]=0.50M, [Al 3+ ]=1.50M E = E o - (0.0592/n)log(Q) (The Nernst Equation) E cell = 0.48V The Nernst Equation (example)

33. Write half-reaction Oxidation : 2Al → 2Al 3+ + 6e - Reduction : 3Mn 2+ + 6e - → 3Mn n=6 (plug in) The Nernst Equation (example)

34. The cell voltage decreases slightly because of the nonstandard concentrations. The reactant concentration is lower than 1.0M and the product concentration is higher than 1.0M. E cell is less than E o cell (E cell =0.48v<E o cell =0.47) More info

35. The potential calculated is the maximum potential before any current flow has occurred. Cell discharges current flows from anode to cathode concentration will change, E cell will change the cell will spontaneously discharge until it reaches equilibrium. Q = K(The equilibrium constant) and E cell = 0 More info

36. A “dead” battery is one in which the cell reaction has reached equilibrium no longer any chemical driving force. At equilibrium the components in the two cell compartments have the same free energy ∆ G = 0 for the cell reaction at the equilibrium concentrations No longer has ability to work More info

37. Describe the cell based on the following half-reactions: Ex 17.7 The Nernst Equation

38. The cell contains components at concentrations other than 1 M. Use Nernst equation n=2 two electrons are transferred Calc. Q Solution x2

39. Solution

40. ● Use the sencitivity to measure potentials to determine the concentration of an ion. (Ex: pH meter) ● Electrodes that are sensitive to the concentration of a particular ion are called ion selective electrodes. Ion-Selective Electrodes

41. A cell at equilibrium: E cell = 0 and Q = K E cell = E o - (0.0591) log(Q) n 0=E o -(0.0591) log(K) n or log(K)= nE o at 25 o C (0.0591) Calculation of Equilibrium Constants for the Redox Reactions

42. half reactions: S 4 O 6 2- + 2e - → 2S 2 O 3 2- E o = 0.17 V Cr 3+ + e - → Cr 2+ E o = -0.50 V 1.Balance the redox reaction and calculate E o and K (at 25 o C) 2. Calculate K log(K) = nE o 0.0591 Warm up – 3/27/15

43. For the oxidation-reduction reaction S 4 O 6 2- (aq) + Cr 2+ (aq) → Cr 3+ (aq) + S 2 O 3 2- (aq) half reactions: S 4 O 6 2- + 2e - → 2S 2 O 3 2- E o = 0.17 V Cr 3+ + e - → Cr 2+ E o = -0.50 V Balance the redox reaction and calculate E o and K (at 25 o C) Ex 17.8 Equilibrium Constants from Cell Potentials

44. 2 moles of electrons are transferred log(K) = nE o = 2(0.67) = 22.6 0.0591 0.0591 K = 10 22.6 = 4x10 22 large equilibrium is not unusual for a redox reaction Solution

45. ● electrolytic cell: uses electrical energy to produce chemical change ● electrolysis: involves forcing a current through a cell to produce a chemical change for which the cell potential is negative ● nonspontaneous chemical reaction 17.7 Electrolysis

46. ● examples: charging a battery, producing aluminum metal, and chrome plating ● (a) Galvanic cell ● (b) Electrolytic (opposite direction)

47. ● external power source forcing electrons through the cell in the opposite direction ● this requires an external potential greater than 1.10V in opposition to the natural cell potential Electrolytic cells

48. ● Electron flow is opposite, the anode and cathode are reversed between (a) and (b) ● ion flow through the salt bridge is opposite in the two cells ● stoichiometry of electrolytic processes: how much chemical change occurs with the flow of a given current for a specified time Electrolytic cells

49. Determine the mass of copper is plated out when a current of 10.0 amps (an ampere [amp], abbreviated A, is 1 coulomb of charge per second) is passed for 30.0 minutes through a solution containing Cu 2+. Electrolytic cells calculation

50. ● Plating means depositing the neutral metal on the electrode by reducing the metal ions in solution. ● Each Cu 2+ ion requires 2 electrons to become an atom of copper metal. Cu 2+ (aq) + 2e - → Cu(s) ● This reduction process will occur at the cathode of the electrolytic cell Electrolytic cells calculation

51. 1. Since an amp is a coulomb of charge per second, multiply the current by the time in seconds to obtain the total coulombs of charge passed into the Cu 2+ solution at the cathode:

52. 2. Since 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulumbs, we can calculate the number of moles This means that 0.187 mole of electrons flowed into the Cu 2+ solution.

53. 3. Each Cu 2+ ion requires two electrons to become a copper atom. Thus each mole of electrons produces ½ mole of copper metal:

54. 4. We now know the moles of copper,etal plated onto the cathode, and we can calculate the mass of copper formed:

55. How long must a current of 5.00 A be applied to a solution of Ag + to produce 10.5 g silver metal? Plan? Ex 17.9 Electroplating

56. Metal plating (electroplating) electrolysis used to coat a material with a layer of the metal. Object to be plated is placed in the cathode. The metal used for plating is placed at the anode. https://www.yo utube.com/wat ch?v=udItd7rjM Ug

57. Metal plating Cathode Anodereduction Deposit metals

58. Hydrogren and oxygen combine spontaneously to form water decreasing in free energy to run fuel cell to produce electricity. The reverse process, which is nonspontaneous by electrolysis. Electrolysis of Water

59. Anode w/1 M H + and cathode w/1 M OH - Pure water - [H + ] = [OH - ] = 10 -7 M potential for the overall process is -1.23V Electrolysis of Water

60. A small amount of a soluble salt causes evolution of bubbles of hydrogen and oxygen. Pure water has negligible ions, so it doesn’t show electrolysis

61. More info If phenolphthalein indicator was put in both sides, what color will the: Anode turns  Produces H + Cathode turns  Produces OH - No change 2H 2 O (l)  O 2 + 4H + + 4 e – Pink 4H 2 O (l) + 4 e –  2H 2 + 4OH -

62. Example: The solution contains Cu 2+, Ag +, Zn 2+. Which order will the metals be plated out onto the cathode? Ag + + e - → Ag E = 0.80V Cu 2+ + 2e - → Cu E = 0.34V Zn 2+ + 2e - → Zn E = -0.76V Electrolysis of MIxtures of Ions

63. More positive the E value, the more the reaction has a tendency Ag + + e - → Ag E = 0.80V Cu 2+ + 2e - → Cu E = 0.34V Zn 2+ + 2e - → Zn E = -0.76V Ag + >Cu 2+ >Zn 2+ Electrolysis of MIxtures of Ions

64. A n acidic solution contains the ions Ce 4+ VO 2 + and Fe 3+. Give the order of oxidizing ability of these species and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage. Ex 17.10 Relative Oxidizing Abilities

65. Order of oxidizing ability Ce 4+ > VO 2 + > Fe 3+ Ce 4+ will be reduced at the lowest voltage Ex 17.10 Relative Oxidizing Abilities

66. Predict the products on cathode and anode Major species Na +, Cl -, H 2 O At the anode: Which is possible to produce? 2Cl - → Cl 2 + 2e - -E = -1.36V 2H 2 O → O 2 + 4h + + 4e -E = -1.23V Water has positive potential, O 2 produced at the anode H 2 O > Cl -. Electrolysis of NaCl solution

67. The voltage is increased, the Cl - ion is the first to be oxidized. Why? A much higher potential is required to oxidize water. (easier to oxidize H 2 O than Cl 2 ) Need higher potential to oxidize water (overvoltage) than Cl 2 Therefore, chlorine is produced first at lower potential. Electrolysis of NaCl solution

68. E values must be used in predicting the actual order of oxidization or reduction of spcies in an electrolytic cell.

69. An unknown metal M is electrolyzed. It took 74.1 seconds for a current of 2.00 amp to plate out 0.107g of the metal from a solution containing M(NO 3 ) 3. Identify the metal. Warm up – 3/30/15

70. Ni2+ (aq) + 2e- → Ni (s)E° = –0.25 V Zn2+ (aq) + 2e- → Zn (s) E° = -0.76 V Can you make a Galvanic cell? If yes, what would be the best metal for anode? What is the cell potential? Warm up – 4/1/15